Suppose we are in the complex case \(|\alpha | < 2\) , and let \(\phi _+ := \arg z_+ \in (0, \pi )\) . On the unit circle \(|z| = 1\) , the function \(\Re S(z) - \Re S(z_+)\)  is zero, with increasing value towards the interior of the unit circle if we're on the left of the vertical line connecting \(z_\pm \) , and increasing towards the exterior if we're on the right of said vertical line. Recall that in \(K_{sp/o}\) , the \(z\)  contour is on the outside of the \(w\)  contour, and we can take without loss of generality the \(z\)  contour outside of the unit circle and the \(w\)  contour on the inside. We then pass both the \(z\)  contour and the \(w\)  contour through the two critical points \(z_\pm \)  in the direction of descent for \(z\)  and ascent for \(w\)  so that in the end, on the right of the vertical line connecting \(z_\pm \) , the \(w\)  contour is outside the unit circle and the \(z\)  contour inside — on the left of said vertical line nothing changes. The new integrals will converge exponentially fast to zero but in exchanging the contours we will have picked up some residues. Due to equations (REF ) and (REF ), there are two types of residues. The first is the classical one of [1]} or of [2]} at \(w z^{-1} = 1\)  or \(z=w\)  — corresponding to the first summand from (REF ) and (REF ) — and the contribution here is a single contour integral over an arc connecting \(z_+\)  to \(z_-\) , oriented bottom to top and passing to the right of 0 — let us call this contour \(\gamma \) . This is the discrete sine part, the integral in question converging to
\( \frac{1}{2 \pi \mathrm {i}} \int _{\gamma } w^{\mathsf {b}- \mathsf {a}- 1} \mathrm {d}w = \frac{\sin (\phi _+ (\mathsf {b}-\mathsf {a}))}{\pi (\mathsf {b}-\mathsf {a})} := K_{sine}(\mathsf {a}, \mathsf {b}).\) 
